# Lower bound of ITM Calls when computing Implied Volatility

Assuming the Black Scholes model and pricing formula of a European call option. Then, if the call is ITM, i.e. if $ln(\frac{S}{K})>0$, the $d_1$-term will go towards infinity as $\sigma$ goes to zero. This also implies that the $d_2$-term will go to infinity and the normal cdfs will both approach 1. This creates a lower bound $S-e^{-r(T-t)}K$ for the option price.

Now let's assume I want to compute the implied volatility for the ITM call option, but the price of call is smaller than lower bound of the B.S. pricing formula. Then the equation I'm trying to solve for the IV does not have a solution. Precisely this is happening as I'm trying to compute the IV of deep ITM calls. However, usually one talks about the volatility smile, where deep ITM calls has larger volatility than ATM calls. Is there any reasonable interpretation of this?

Just as a final note, since you are talking about deep ITM calls (or OTM puts), the volatility can go up due to the fact that the prices are quantized. That is, all OTM puts from some point will cost exactly \$0.01, simply because there is nothing below that. Obviously, the implied volatility for higher strike will be higher in such case, with this price fixed and you'll see it going straight up from some point. This can be mistaken for the genuine volatility smile, while it is not it - in theory, option price should go below \$0.01 to fractions and the implied volatility would be completely different.